Question

Data Collection Barometric pressure, Pear (mm Hg) 737,3 Temperature and Volume Measurements Table 1. Temperature of water and volume of air bubble Temperature of water (°C) 78.4 Volume of air bubble (mL) 8.82 Reading 1 Reading 2 75.0 7.49 70.0 6.33 65.0 5.34 60.5 4.96 Reading 3 Reading 4 Reading 5 Reading 6 Reading 7 Reading 8 55.0 4.52 50.1 4.39 3.9 3.12 (31 pts) Calculations Use the temperatures and volumes collected in Table 1 to complete the calcualtions for Readings 1-7. Note: Since we assume that Pear = Pair at 3.9 °C, we omit the calculations for that temperature because Pwater would equal O, and in 0 is undefined. Table view List view

Please help!

Answer 1

ANSWER:

To fill the table we need the following equations (there is an example of calculations with the data of Reading 1):

• Temperature of water in Kelvin

TK = T. + 273.15

for Reading 1

TK = 78.4°C + 273.15 = 351.55 K

• 1 /T (for reading 1)

$\frac{1}{T_{K}}=\frac{1}{351.55\, K}=2.84x10^{-3}\, K^{-1}$

• Pressure of air (mmHg)

$P_{ReadingX}V_{ReadingX}=P_{Reading8}V_{Reading8}$

Pair R.X Vair air RX Pair R. Vair R.8 Pair Rx= Pair R.8 Vair R.8 Vair RX

for Reading 8, P_air = P_Bar = 737.3 mmHg, then

Pair R.X 737.3 mm Hg x 3.12 mL Vair RX

For Reading 1, we have

$P_{air\, R.1}=\frac{737.3\, mmHg\times 3.12\, mL }{V_{air\, R.1}}=\frac{737.3\, mmHg\times 3.12\, mL }{8.82\, mL}=260.81\, mmHg$

• Pressure of water (mmHg)

For Readings 1-7, we have

$P_{Bar}=P_{air}+P_{water}\, \, \, \, \, \Rightarrow \, \, \, \, \, P_{water}=P_{Bar}-P_{air}$

for Reading 1:

• Pressure of water (atm)

$P_{water} (atm)=P_{water} (mmHg)\times \frac{1\, atm}{760\, mmHg}$

for Reading 1:

$P_{water} (atm)=476.49\, mmHg\times \frac{1\, atm}{760\, mmHg}=0.627\, atm$

Now, filling the table

	Twater (ºC)	Twater (K)	1/Twater (K-1)	Vair (mL)	Pair (mmHg)	Pwater (mmHg)	Pwater (atm)	ln Pwater
Reading 1	78.4	351.55	2.84E-03	8.82	260.814	476.49	0.627	-0.467
Reading 2	75.0	348.15	2.87E-03	7.49	307.126	430.17	0.566	-0.569
Reading 3	70.0	343.15	2.91E-03	6.33	363.409	373.89	0.492	-0.709
Reading 4	65.0	338.15	2.96E-03	5.34	430.782	306.52	0.403	-0.908
Reading 5	60.5	333.65	3.00E-03	4.96	463.785	273.51	0.360	-1.022
Reading 6	55.0	328.15	3.05E-03	4.52	508.933	228.37	0.300	-1.202
Reading 7	50.1	323.25	3.09E-03	4.39	524.004	213.30	0.281	-1.271

With this dat we made the plot of ln P (y-axis) vs. 1/T (x-axis) and we add the best-fit line equation:

In P vs 1/T 0.00 -0.20 -0.40 -0.60 In Pwater -0.80 -1.00 Y = -3654.6% +9.9273 R=0.9974 -1.20 -1.40 2.80E-03 2.85E-03 2.90E-03 2.95E-03 3.00E-03 3.05E-03 3.10E-03 3.15E-03 1/T water (K)

• To get the best correlation factor (R²), we do not use the data of Reading 7 for the linear regression.

According the grap, the best-fit line equation is

$y=-3654.6x+9.9273$

or

$ln\, P_{water}=-3654.6\left ( \frac{1}{T_{water}} \right )+9.9273$

This equation has the same form that the following equation:

$ln\, P_{water}=- \frac{\Delta H^{\circ}_{vapor}}{R} \left ( \frac{1}{T_{water}} \right )+\frac{\Delta S^{\circ}_{vapor}}{R}$

Then, we can get ∆Hº_vapor and ∆Sº_vapor:

• ∆Hº_vapor

$slope=- \frac{\Delta H^{\circ}_{vapor}}{R}=-3654.6\, K$

$\frac{\Delta H^{\circ}_{vapor}}{R}=3654.6\, K$

$\Delta H^{\circ}_{vapor}=3654.6\, K\times 8.314\, \frac{J}{mol.K}=30384\, \frac{J}{mol}=\mathbf{30.38\, \frac{kJ}{mol}}$

• ∆Sº_vapor

$y-intercept= \frac{\Delta S^{\circ}_{vapor}}{R}=9.9273$

$\Delta S^{\circ}_{vapor}=9.9273 \times 8.314\, \frac{J}{mol.K}=\mathbf{82.536\, \frac{J}{mol.K}}$

Now, we can calculate the value of ∆Gº_vapor at 298 K

$\Delta G^{\circ}_{vapor}=\Delta H^{\circ}_{vapor}-T\Delta S^{\circ}_{vapor}$

$\Delta G^{\circ}_{vapor}=30384\, \frac{J}{mol}-298\, K\times 82.536\, \frac{J}{mol.K}$

$\Delta G^{\circ}_{vapor}=5788\, \frac{J}{mol}=\mathbf{5.79\, \frac{kJ}{mol}}$

The normal boilling point could be calculated with the line equation from graph:

$ln\, P=-3654.6\left ( \frac{1}{T} \right )+9.9273$

for the normal boilling point, the pressure is 1 atm, then

$ln\, 1=-3654.6\left ( \frac{1}{T} \right )+9.9273$

$\frac{1}{T} =\frac{ln\, 1-9.9273}{-3654.6}=2.72x10^{-3}\, K^{-1}$

$T=\frac{1}{2.72x10^{-3}\, K^{-1}} =368.14\, K$

$T=368.14\, K-273.15=\mathbf{94.99\, ^{\circ}C}$

The table 3 could be filled as

	Experimental	Literature	% error
∆Hºvapor for water (kJ/mol)	30.38	40.66	25.28 %
∆Sºvapor for water (J/mol)	82.536	108.96	24.25 %
∆Gºvapor for water at 298 K(kJ/mol)	5.79	8.19	29.30 %
Normal boilling point for water in ºC	94.99	100	5.01 %

The percent error is calculated as

$\%\, error=\frac{\left | Experimental-Literature \right |}{Experimental}\times 100\%$